# Prove convergence of $x_i$, where $x_1 = 1$ and $x_{n+1}=(\sqrt{2})^{x_n}$

I need to prove that the sequence $\{x_n\}_{n=1}^\infty$ converges, where $x_1=1$ and $$x_{n+1}=(\sqrt{2})^{x_n}$$ The only way I can see to progress is to use the monotone bounded sequence theorem. (Nothing else makes sense to me.) My intuition tells me the sequence should be bounded above by $2$, and that $x_{n+1}>x_n$, but I can't work out what to do to prove those.

• At the end, you want $x_{n+1}\gt x_n$. We haven't heard about $a$ before and you have the sense of the inequality wrong. You are correct that it is bounded above by $2$. Assume $x_n \lt 2$ and use the monotonicity of $\sqrt 2^x$ to conclude $x_{n+1} \lt 2$ Apr 17, 2018 at 22:42
• sorry, dont know why i changed to $a_n$, I'll fix that. Apr 17, 2018 at 23:07

If $x_n < 2$, then
$$x_{n + 1} = \sqrt{2}^{x_n} < \sqrt{2}^2 = 2.$$
Furthermore, the function $x \mapsto 2^{x/2}$ is a strictly increasing function (its derivative is $2^{x/2} \cdot \ln \sqrt 2 > 0$, for example), and so $x_{n + 1} > x_n$.